quantum field theory - photon propagator

Just multiply it out in Fourier, where $partial = ik$:

$$bigl[ (-k^2 +m^2) g^{munu} + k^mu k^nubigr] frac{ -g_{nulambda} + m^{-2} k_nu k_lambda }{k^2 - m^2} = frac1{k^2 - m^2} bigl( - (-k^2 + m^2) delta^mu_lambda + (-k^2 + m^2) m^{-2} k^mu k_lambda - k^mu k_lambda + k^mu k^2 k_lambda m^{-2} bigr) = delta^mu_lambda$$

which is a function of $k$. But converting back to position space, $1(k) = delta(x)$.

This proves that $D$ is a solution. To be the solution, you usually have to impose boundary conditions, etc. In this case, there are no solutions to $bigl[ (-k^2 +m^2) g^{munu} + k^mu k^nubigr] f_nu = 0$: the corresponding equation in Fourier is $(-k^2 + m^2) g^{munu} + k^mu k^nu = 0$, and contracting with $g_{mu nu}$ gives $0 = d(-k^2 + m^2) + k^2 = dm^2 - (d-1)k^2$, where $d$ is the dimension of spacetime, so $k^2 = frac{d}{d-1}m^2$, but $-frac1{d-1}m^2 g^{munu} + k^mu k^nu$ cannot equal $0$, as $k^mu k^nu$ cannot be an invertible matrix. So $D$ is the only solution.

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